For more information, please see full course syllabus of Pre Algebra

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For more information, please see full course syllabus of Pre Algebra

For more information, please see full course syllabus of Pre Algebra

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Related Books

### Solving Equations with Variables on Both Sides

- To solve equations with variables on both sides, bring all the variable terms to one side of the equation. Use the Distributive Property, if necessary, collect like terms, and simplify.
- To solve equations with variables on both sides, simplify as much as you can on both sides before you apply inverse operations to isolate the variable.
- After finding the solution, substitute to check your answer. In real life situations, ask yourself if the answer is reasonable.

### Solving Equations with Variables on Both Sides

5x + 15 = 8x. Solve for x.

- 5x − 8x = − 15
- − 3x = − 15

x = 3

− 8 − 4x = 10 + 3x. Solve for x.

- − 8 − 10 = 3x + 7x
- − 18 = 10x
- x = − [18/10]

x = − [9/5]

4x + 5 = 3(2x + 1). Solve for x.

- 4x + 5 = 6x + 3
- 4x − 6x = 3 − 5
- − 2x = − 2

x = 1

11 − 5x + 3 = − 3x − 10. Solve for x.

- − 5x + 3x = − 10 − 11 − 3
- − 2x = − 24

x = 12

3.6x = 4(x − 2). Solve for x.

- 3.6x = 4x − 8
- 3.6x − 4x = − 84
- − 0.4x = − 8

x = 20

35 + 7x = 24 + x + 1. Solve for x.

- 7x − x = 24 + 1 − 35
- 6x = 25 − 35
- 6x = − 104

x = − [5/3]

14.5 + 5x = − 3.3x − 10.4. Solve for x.

- 5x + 3.3x = − 10.4 − 14.5
- 8.3x = − 24.9

x = − 3

− 3(5x − 2) = − 18x + 10x. Solve for x.

- − 15x + 6 = − 8x
- − 15x + 8x = − 6
- − 7x = − 6

x = [6/7]

Two boats are taking the same path to a destination. Ship A travels at 25 mi/hr, while ship B travels at 30 mi/hr and leaves 30 min after ship A does. How long after ship A leaves will ship B catch up to ship A?

- Let t = time that ship A travels before ship B catches up.
- distance
_{ship A}= distance_{ship B}

distance = rate × time - 25 mi/hr ×t = 30 mi/hr × (t - 0.5)
- 25t = 30t − 15
- 25t − 30t = − 15
- − 5t = − 15

t = 3 hours

Lisa runs [1/8] mi/min. George runs [1/10] mi/min. Lisa starts running 5 min after George starts. How long after George starts running have they run the same distance?

- Let t = number of minutes
- distance
_{Lisa}= distance_{George}

distance = rate × time - [1/8](t − 5) = [1/10]t
- [1/8]t − [5/8] = [1/10]t
- [1/8]t − [1/10]t = [5/8]
- [5/40]t − [4/40]t = [25/40]
- [1/40]t = [25/40]

t = 25 min

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Solving Equations with Variables on Both Sides

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- What You'll Learn and Why
- Vocabulary
- Variables on Both Sides
- Using the Distributive Property
- Using the Distributive Property
- Extra Example 1: Solving Equations with Variables on Both Sides
- Extra Example 2: Solving Equations with Variables on Both Sides
- Extra Example 3: Solving Equations with Variables on Both Sides
- Extra Example 4: Cost of Renting Video

- Intro 0:00
- What You'll Learn and Why 0:07
- Topics Overview
- Vocabulary 0:26
- Term
- Like Terms
- Variables on Both Sides 1:10
- Example: 3x + 24 = 9x
- Example: -7 - 3x = 1 + 5x
- Using the Distributive Property 4:01
- Example: Height of Two Plants
- Using the Distributive Property 9:01
- Example: Running Laps
- Extra Example 1: Solving Equations with Variables on Both Sides 11:59
- Extra Example 2: Solving Equations with Variables on Both Sides 12:46
- Extra Example 3: Solving Equations with Variables on Both Sides 14:18
- Extra Example 4: Cost of Renting Video 15:24

0 answers

Post by sherman boey on August 22, 2014

better check the answers for this topic and the previous alot of wrong answers this makes new students confuse..